Problem of the Week #131 - September 29th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Find the shortest distance from the point $S(-8,-10,-8)$ to the line defined parametrically by $x=-4+t$, $y=-9+2t$, $z=-3+2t$.

-----Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Kiwi. You can find their answer below.

[sp]d=\sqrt{(S_x-x)^2+(S_y-y)^2(S_z-z)^2}

\therefore d=\sqrt{(-8+4-t)^2+(-10+9-2t)^2(-8+3-2t)^2}

\therefore d=\sqrt{(t+4)^2+(2t+1)^2+(2t+5)^2}

\therefore \frac {dd}{dt}=\frac 12 ((t+4)^2+(2t+1)^2+(2t+5)^2)^{-\frac 12}[2(t+4)+2(2t+1)2+2(2t+5)2]

\therefore \frac {dd}{dt}=\frac 12 (t^2+8t+16+4t^2+4t+1+4t^2+20t+25)^{-\frac 12}[2t+8+8t+4+8t+20]

\therefore \frac {dd}{dt}=\frac 12 (9t^2+32t+42)^{-\frac 12}[18t+32]

Now at the point of minimum the derivative equals zero so:

18t+32=0

\therefore t=-\frac{16}{9}

So
x=-4+t=-4-\frac{16}{9}=-\frac{52}{9}

y=-9+2t=-9-\frac{32}{9}=\frac{-113}{9}

z=-3+2t=-3-\frac{32}{9}=-\frac{-59}{9}

And

d=\sqrt{(-8+\frac{52}{9})^2+(-10+\frac{113}{9})^2(-8+\frac{59}{9})^2}

\therefore d=\sqrt{\frac{400+529+169}{81}}

\therefore d=\frac 13 \sqrt{122}[/sp]